Every object that has mass attracts every other object with a gravitational force: It has been proven that the size of the

2 answers:

8 0

Answer;

- This statement about matter and its behavior is best classified as a Law.
-It is the Law of Universal Gravitation.

Explanation;
-The Law of Universal Gravitation states that every point mass attracts every other point mass in the universe by a force pointing in a straight line between the centers-of-mass of both points, and this force is proportional to the masses of the objects and inversely proportional to their separation.This attractive force always points inward, from one point to the other.
-The Law applies to all objects with masses, big or small. Two big objects can be considered as point-like masses, if the distance between them is very large compared to their sizes or if they are spherically symmetric.

denpristay [2]3 years ago

7 0

Answer:

this is a law because it is a constant fact of nature

Explanation:

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A rigid tank initially contains 1.4 kg saturated liquid water at 200◦C. At this state, 25 percent of the volume is occupied by w

rjkz [21]

Answer:

(a) Volume of the tank is $6.47\times 10^{- 3}\ m^{3}$

(b) Temperature is $371^{\circ}C$

Pressure is 21.3 kPa

Solution:

As per the question:

Mass of liquid in the tank, m = 1.4 kg

Temperature, T = $200^{\circ}C$

Volume occupied by water, V = 25% $V_{t}$ = 0.25 $V_{t}$

Volume occupied by air, V' = 75% $V_{t}$

where

$V_{t}$ = Volume of tank

Now,

(a) In order to calculate the volume of the tank, we make use of the steam table for specific volume at as given temperature:

At $200^{\circ}C$ , the specific volume, v = 0.001157 $m^{3}/kg$

At $200^{\circ}C$ , the internal energy, u = 850.46 kJ/kg

Now, Volume of water, V = mv = $1.4\times 0.001157 = 1.62\times 10^{- 3} m^{3}$

Thus

$V = 0.25V_{t}$

$V_{t} = \frac{1.62\times 10^{- 3}}{0.25} = 6.47\times 10^{- 3}\ m^{3}$

(b) For the final temperature and pressure, we calculate the specific volume, v' and then find the corresponding temperature and pressure from the steam table:

v' = $\frac{V_{t}}{m} = \frac{6.47\times 10^{- 3}}{1.4} = 4.63\times 10^{- 3}\ m^{3}/kg$